# Construct group isomorphism $\alpha : \mathbb{Z}^n \rightarrow \mathbb{Z}^n$ s.t. $g=f \circ \alpha$?

Suppose we have surjective group morphisms $$f: \mathbb{Z}^n\rightarrow A \qquad g:\mathbb{Z}^n\rightarrow A.$$

How do I construct a group isomorphism $\alpha:\mathbb{Z}^n \rightarrow \mathbb{Z}^n$ such that $g=f \circ \alpha$ ?

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@Nadori Yikes, I didn't see that you wanted an isomorphism. I guess we'll have to be more careful. Is it so clear that you can always do this? –  Dylan Moreland Feb 26 '12 at 17:35
You can't: $n=1, A=\mathbb Z/8, g=can, f=3\cdot can$ –  Georges Elencwajg Feb 26 '12 at 17:38
@GeorgesElencwajg I think this is the true "answer" to the question. Very nice! –  Dylan Moreland Feb 26 '12 at 17:41
Thanks, @Dylan. –  Georges Elencwajg Feb 26 '12 at 17:45

Apparently I don't have the privilege of adding a comment yet, but I would like to point out that Dylan's answer is correct since a homomorphism on a free abelian group is determined by where it sends the basis. HOWEVER the original problem has no general solution since in general $\alpha$ will not be an isomorphism.

For example, if $n=1$, $A=\mathbb{Z}/5\mathbb{Z}$, $f(1)=[1]$ and $g(1)=[2]$ (surjective since $2$ and $5$ are coprime), then there is no isomorphism $\alpha:\mathbb{Z}\rightarrow\mathbb{Z}$ such that $g=f\circ\alpha$

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That $g$ is not surjective. –  Thomas Andrews Feb 26 '12 at 17:53
$g$ is not surjective ; your counter example doesn't work. That's why I downvoted. I'll take it off if you change something to your answer. –  Patrick Da Silva Feb 26 '12 at 18:18
Fixed, sorry about that –  you Feb 26 '12 at 18:27
@you : Yes, now that's better. +1! In other words, there's an isomorphism "inside" (i.e. from $A$ to $A$) but not "outside" (from $\mathbb Z^n$ to $\mathbb Z^n$) that completes the diagram. Impressive! You can write $g = \alpha \circ f$ but not $g = f \circ \alpha$. Good eye. –  Patrick Da Silva Feb 26 '12 at 18:58

This was way too long for a comment so I'm posting this here. Don't consider this as a complete answer ; I am saying no such thing. I am only saying that the problem is not as general as it seems.

$f : \mathbb Z^n \to A$ is surjective hence $\mathbb Z^n / \mathrm{Ker} \, f \cong A$. For the same reasons, $\mathbb Z^n/ \mathrm{Ker} \, g \cong A$, hence $$\mathbb Z^n / \mathrm{Ker} \, f \cong \mathbb Z^n / \mathrm{Ker} \, g.$$ Now all this being said, we have the following commutative diagram : $$\begin{matrix} \mathbb Z^n & \overset{\alpha}{\longleftarrow} & \mathbb Z^n \\\ \downarrow f & & \downarrow g \\\ \mathbb Z^n/ \mathrm{Ker} f & \overset{\varphi_1^{-1} \circ i \circ \varphi_2}{\longleftarrow}& \mathbb Z^n/ \mathrm{Ker} g \\\ \downarrow \varphi_1 & & \downarrow \varphi_2 \\\ A & \overset{i}{\longleftarrow} & A \\\ \end{matrix}$$ where $\varphi_1$, $\varphi_2$ are isomorphisms and $i$ is the inclusion map. The question is whether $\alpha$ fits in there as an isomorphism (i.e. does it exist). Well, we can reduce ourselves to the study of the map $\psi = i \circ \varphi_2 \circ g$ and look at the diagram $$\begin{matrix} \mathbb Z^n & \overset{\psi}{\longrightarrow} & A \\\ & & \uparrow \varphi_1 \\\ & & \mathbb Z^n / \mathrm{Ker} f\\\ \end{matrix}$$ and ask ourselves if this diagram commutes. Since $\varphi_1$ is an isomorphism it is injective and $\psi$ is surjective, therefore we can complete this diagram with a unique morphism $\Phi$ such that $\psi = \varphi_1 \circ \Phi$. Now I don't know how to TeX this but that gives us $\Phi$ going from the topright $\mathbb Z^n$ to the middle left $\mathbb Z^n / \mathrm{Ker} f$, and I've shown that this map is unique. All we need to do now is complete the diagram $$\begin{matrix} \mathbb Z^n & \overset{\Phi}{\longrightarrow} & \mathbb Z^n / \mathrm{Ker} f \\\ & & \uparrow f \\\ & & \mathbb Z^n \\\ \end{matrix}$$ so that in other words, I've reduced the problem for arbitrary $A$ to the problem of dealing with some quotient of $\mathbb Z^n$ ; given $f : \mathbb Z^n \to \mathbb Z^n$ and $\Phi : \mathbb Z^n \to \mathbb Z^n / \mathrm{Ker} f$, one completes this diagram if and only if one completes the diagram above for arbitrary $f$,$g$. Since there is no obvious reason why this diagram should be completed I expect counter examples.

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